A008279 -id:A008279 - cp's OEIS frontend

A010843 Incomplete Gamma Function at -3.

1, -2, 5, -12, 33, -78, 261, -360, 3681, 13446, 193509, 1951452, 23948865, 309740922, 4341155877, 65102989248, 1041690874689, 17708615729550, 318755470552389, 6056352778233924, 121127059051462881, 2543668229620367298

Offset: 1

Simon Plouffe

sign

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 262.

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 262.

Cf. A008279 A008279 A089258.

a := n -> n!*add(((-3)^(k)/k!), k=0..n): seq(a(n), n=0..21); # Zerinvary Lajos, Jun 22 2007
seq(simplify(KummerU(-n, -n, -3)), n = 0..21); # Peter Luschny, May 10 2022

Table[ Gamma[ n, -3 ]*E^(-3), {n, 1, 24} ] (* corrected by Peter Luschny, Oct 17 2012 *)
a[n_] := (-1)^n x D[1/x Exp[x], {x, n}] x^n Exp[-x]
Table[a[n] /. x -> 3, {n, 0, 20}] (* Gerry Martens , May 05 2016 *)

a(n)=if(n<0,0,n!*polcoeff(exp(-3*x+x*O(x^n))/(1-x),n)) /*  Michael Somos, Mar 06 2004 */

a(n)=local(A,p);if(n<1,n==0,A=matrix(n,n,i,j,1-3*(i==j));sum(i=1,n!,if(p=numtoperm(n,i),prod(j=1,n,A[j,p[j]])))) /* Michael Somos, Mar 06 2004 */

@CachedFunction
def A010843(n):
    if (n) == 1 : return 1
    return (n-1)*A010843(n-1)+(-3)^(n-1)
[A010843(i) for i in (1..22)]    # Peter Luschny, Oct 17 2012

E.g.f.: exp(-3x)/(1-x). - Michael Somos, Mar 06 2004
a(0) = 1 and for n>0, a(n) is the permanent of the n X n matrix with -2's on the diagonal and 1's elsewhere. a(n) = Sum(k=0..n, A008290(n, k)*(-2)^k ). a(n) = Sum(k=0..n, A008279(n, k)*(-3)^(n-k) ). - Philippe Deléham, Dec 15 2003
G.f.: hypergeom([1,1],[],x/(1+3*x))/(1+3*x). - Mark van Hoeij, Nov 08 2011
E.g.f.: 1/E(0) where E(k)=1-x/(1-3/(3-(k+1)/E(k+1))); (continued fraction). - Sergei N. Gladkovskii, Sep 13 2012
G.f.:  1/Q(0), where Q(k)= 1 + 3*x - x*(k+1)/(1-x*(k+1)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, Apr 18 2013
G.f.: 1/Q(0), where Q(k) = 1 - x*(2*k-2) - x^2*(k+1)^2/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Sep 30 2013
a(n) ~ n! * exp(-3). - Vaclav Kotesovec, Oct 08 2013
a(n) = (-3)^(n-1)*hypergeom([1, 1-n], [], 1/3). - Vladimir Reshetnikov, Oct 18 2015
a(n) = KummerU(-n, -n, -3). - Peter Luschny, May 10 2022

A072374 a(1) = 1; a(n) = 1 + Sum_{i=1..n} Product_{j=i..2*i-1} (n-j).

Original entry on oeis.org

1, 2, 3, 6, 11, 24, 51, 122, 291, 756, 1979, 5526, 15627, 46496, 140451, 442194, 1414931, 4687212, 15785451, 54764846, 193129659, 698978136, 2570480147, 9672977706, 36967490691, 144232455524, 571177352091, 2304843053382, 9434493132011, 39289892366736

Offset: 1

N. J. A. Sloane, Jul 19 2002

nonn

A122852 is another version of the same sequence. - R. J. Mathar, Jun 18 2008

Vincenzo Librandi, Table of n, a(n) for n = 1..200

Table[Sum[Binomial[n-k,k]*k!,{k,0,Floor[n/2]}],{n,1,20}] (* Vaclav Kotesovec, Feb 08 2014 *)

The sequence 1, 1, 2, 3, .. has a(n)=sum{k=0..floor(n/2), C(n-k, k)k!} (diagonal sums of permutation triangle A008279). - Paul Barry, May 12 2004
Recurrence: 2*a(n) = 3*a(n-1) + (n-1)*a(n-2) - (n-1)*a(n-3). - Vaclav Kotesovec, Feb 08 2014
a(n) ~ sqrt(Pi) * exp(sqrt(n/2) - n/2 + 1/8) * n^((n+1)/2) / 2^(n/2+1) * (1 + 37/(48*sqrt(2*n))). - Vaclav Kotesovec, Feb 08 2014

A089504 A generalization of triangle A071951 (Legendre-Stirling).

Original entry on oeis.org

1, 6, 1, 36, 30, 1, 216, 756, 90, 1, 1296, 18360, 6156, 210, 1, 7776, 441936, 387720, 31356, 420, 1, 46656, 10614240, 23705136, 4150440, 119556, 756, 1, 279936, 254788416, 1432922400, 521757936, 29257200, 373572, 1260, 1, 1679616

Offset: 1

Wolfdieter Lang, Dec 01 2003

This triangle underlies the array entry A078741 ((3,3)-generalized Stirling2).

For the computation of the column sequences see A089505.

			[1]; [6,1]; [36,30,1]; [216,756,90,1]; ...
a(3,2) = 30 = ((-1)*(3*2*1)^1 + 4*(4*3*2)^1)/3.

R. B. Corcino, K. J. M. Gonzales, M. J. C. Loquias and E. L. Tan, Dually weighted Stirling-type sequences, arXiv preprint arXiv:1302.4694 [math.CO], 2013.
R. B. Corcino, K. J. M. Gonzales, M. J. C. Loquias and E. L. Tan, Dually weighted Stirling-type sequences, Europ. J. Combin., 43, 2015, 55-67.
W. Lang, First 8 rows.

Cf. A071951 (Legendre-Stirling, (2, 2) case).
The column sequences (without leading zeros) are A000400 (powers of 6), A089507, A089513-4, etc.
Cf. A008279, A089505, A089506.

max = 10; f[m_] := 1/Product[1 - FactorialPower[r + 2, 3]*x, {r, 1, m}]; col[m_] := CoefficientList[f[m] + O[x]^(max - m + 1), x]; a[n_, m_] := col[m][[n - m + 1]]; Table[a[n, m], {n, 1, max}, {m, 1, n}] // Flatten (* Jean-François Alcover, Sep 01 2016 *)

G.f. for m-th column sequence (without leading zeros and m>=1) is 1/Product_{r=1..m} 1-fallfac(r+2, 3)*x with fallfac(n, k) := A008279(n, k) (falling factorials).

a(n, m) = Sum_{p=1..m} A089505(m, p)*((p+2)*(p+1)*p)^(n-m))/D(m) if n>=m>=1 else 0; with D(m) := A089506(m).

A090214 Generalized Stirling2 array S_{4,4}(n,k).

Original entry on oeis.org

1, 24, 96, 72, 16, 1, 576, 13824, 50688, 59904, 30024, 7200, 856, 48, 1, 13824, 1714176, 21606912, 76317696, 110160576, 78451200, 30645504, 6976512, 953424, 78400, 3760, 96, 1, 331776, 207028224, 8190885888, 74684104704, 253100173824

Offset: 1

Wolfdieter Lang, Dec 01 2003

The row length sequence for this array is [1,5,9,13,17,...] = A016813(n-1), n >= 1.
The g.f. for the k-th column, (with leading zeros and k >= 4) is G(k,x) = x^ceiling(k/4)*P(k,x)/Product_{p = 4..k} (1 - fallfac(p,4)*x), with fallfac(n,m) := A008279(n,m) (falling factorials) and P(k,x) := Sum_{m = 0..kmax(k)} A090221(k,m)*x^m, k >= 4, with kmax(k) := A057353(k-4)= floor(3*(k-4)/4). For the recurrence of the G(k,x) see A090221.
Codara et al., show that T(n,k) gives the number of k-colorings of the graph nK_4 (the disjoint union of n copies of the complete graph K_4). - Peter Bala, Aug 15 2013

			Table begins
n\k|   4      5      6      7      8     9   10   11   12
= = = = = = = = = = = = = = = = = = = = = = = = = = = = =
1  |   1
2  |  24     96     72     16      1
3  | 576  13824  50688  59904  30024  7200  856   48    1
...

Robert Israel, Table of n, a(n) for n = 1..10011 (rows 1 to 71, flattened)
P. Blasiak, K. A. Penson and A. I. Solomon, The general boson normal ordering problem, arXiv:quant-ph/0402027, 2004.
P. Blasiak, K. A. Penson and A. I. Solomon, The general boson normal ordering problem, Phys. Lett. A 309 (2003) 198-205.
P. Codara, O. M. D’Antona, and P. Hell, A simple combinatorial interpretation of certain generalized Bell and Stirling numbers, arXiv:1308.1700v1 [cs.DM], 2013.
A. Dzhumadildaev and D. Yeliussizov, Path decompositions of digraphs and their applications to Weyl algebra, arXiv preprint arXiv:1408.6764v1 [math.CO], 2014. [Version 1 contained many references to the OEIS, which were removed in Version 2. - _N. J. A. Sloane_, Mar 28 2015]
Askar Dzhumadil’daev and Damir Yeliussizov, Walks, partitions, and normal ordering, Electronic Journal of Combinatorics, 22(4) (2015), #P4.10.
Wolfdieter Lang, First 4 rows.
M. Schork, On the combinatorics of normal ordering bosonic operators and deforming it, J. Phys. A 36 (2003) 4651-4665.

Cf. A090215, A071379 (row sums), A090213 (alternating row sums).

S_{1, 1} = A008277, S_{2, 1} = A008297 (ignoring signs), S_{3, 1} = A035342, S_{2, 2} = A078739, S_{3, 2} = A078740, S_{3, 3} = A078741.

T:= (n,k) -> (-1)^k/k!*add((-1)^p*binomial(k,p)*(p*(p-1)*(p-2)*(p-3))^n,p=4..k):
seq(seq(T(n,k),k=4..4*n),n=1..10); # Robert Israel, Jan 28 2016

a[n_, k_] := (((-1)^k)/k!)*Sum[((-1)^p)*Binomial[k, p]*FactorialPower[p, 4]^n, {p, 4, k}]; Table[a[n, k], {n, 1, 5}, {k, 4, 4*n}] // Flatten (* Jean-François Alcover, Sep 05 2012, updated Jan 28 2016 *)

a(n, k) = (-1)^k/k! * Sum_{p = 4..k} (-1)^p * binomial(k, p) * fallfac(p, 4)^n,  with fallfac(p, 4) := A008279(p, 4) = p*(p - 1)*(p - 2)*(p - 3); 4 <= k <= 4*n, n >= 1, else 0. From eq.(19) with r = 4 of the Blasiak et al. reference.
E^n = Sum_{k = 4..4*n} a(n,k)*x^k*D^k where D is the operator d/dx, and E the operator (x^4)*d^4/dx^4.
The row polynomials R(n,x) are given by the Dobinski-type formula R(n,x) = exp(-x)*Sum_{k >= 0} (k*(k - 1)*(k - 2)*(k - 3))^n*x^k/k!. - Peter Bala, Aug 15 2013

A093964 a(n) = Sum_{k=1..n} kk!C(n,k).

Original entry on oeis.org

0, 1, 6, 33, 196, 1305, 9786, 82201, 767208, 7891281, 88776910, 1085051121, 14322674796, 203121569833, 3080677142466, 49764784609065, 853110593298256, 15469738758475041, 295858753755835158, 5951981987323272001, 125652953065713520020, 2777591594084193600441

Offset: 0

Ralf Stephan, Apr 20 2004

nonn

Limit to which the columns of array A093966 converge.
Number of objects in all permutations of n objects taken 1,2,...,n at a time. Example: a(2)=6 because the permutations of {a,b} taken 1 and 2 at a time are: a,b,ab and ba, containing altogether 1+1+2+2=6 objects. a(n)=Sum(k*A008279(n,k),k=1..n). - Emeric Deutsch, Aug 16 2006
The number of sequences -where each member is an element in a set consisting of n elements- such that the last member is a repetition of a former member. Example: Set of possible members: {l,r}. Sequences such that the last member is a repetition of a former member: l,l; r,r; l,r,l; l,r,r; r,l,l; r,l,r. a(n)=Sum(k*A008279(n,k),k=1..n). [From Franz Fritsche (ff(AT)simple-line.de), Feb 22 2009]
The total number of elements in all ascending runs (including runs of length 1) over all permutations of {1,2,...,n}.  a(2) = 6 because in the permutations [1,2] and [2,1] there are 4 runs of length 1 and 1 run of length 2. a(n) = Sum_{k>=1} A132159(n,k)*k. - Geoffrey Critzer, Feb 24 2014

			G.f. = x + 6*x^2 + 33*x^3 + 196*x^4 + 1305*x^5 + 9786*x^6 + 82201*x^7 + ...

Alois P. Heinz, Table of n, a(n) for n = 0..200

Cf. A000522, A001339, A007526, A008279.

Row n=2 of A210472. - Alois P. Heinz, Jan 23 2013

[0] cat [n le 2 select 6^(n-1) else n*((n+1)*Self(n-1) - (n-1)*Self(n-2))/(n-1): n in [1..30]]; // G. C. Greubel, Dec 29 2021

seq(add(k*n!/(n-k)!,k=1..n),n=0..20); # Emeric Deutsch, Aug 16 2006
# second Maple program:
a:= proc(n) a(n):=`if`(n<2, n, n*((n+1)/(n-1)*a(n-1)-a(n-2))) end:
seq(a(n), n=0..30);  # Alois P. Heinz, Jan 21 2013

nn=21;Range[0,nn]!CoefficientList[Series[D[Exp[y x]/(1-x)^2,y]/.y->1,{x,0,nn}],x] (* Geoffrey Critzer, Feb 24 2014 *)

PARI
```
a(n)=sum(k=1,n,k*k!*binomial(n,k))
```

[factorial(n)*( x*exp(x)/(1-x)^2 ).series(x,n+1).list()[n] for n in (0..30)] # G. C. Greubel, Dec 29 2021

E.g.f.: x*exp(x)/(1-x)^2. - Vladeta Jovovic, Apr 24 2004
a(n) = 1 + (n-1)*floor(e*n!) = 1 + (n-1)*A000522(n) = A000522(n+1) - 2*A000522(n) = A001339(n) - A000522(n). - Henry Bottomley, Dec 22 2008
a(n) = n if n < 2, a(n) = n*((n+1)/(n-1)*a(n-1) - a(n-2)) for n >= 2. - Alois P. Heinz, Jan 21 2013
E.g.f.: x*(1- 12*x/(Q(0)+6*x-3*x^2))/(1-x)^2, where Q(k) = 2*(4*k+1)*(32*k^2+16*k+x^2-6) - x^4*(4*k-1)*(4*k+7)/Q(k+1) ; (continued fraction). - Sergei N. Gladkovskii, Nov 18 2013
G.f.: conjecture: T(0)/x - 1/x, where T(k) = 1 - x^2*(k+1)^2/(x^2*(k+1)^2 - (1 - 2*x*(k+1))*(1 - 2*x*(k+2))/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 18 2013
a(n) = n*a(n-1) + A007526(n), a(0) = 0. - David M. Cerna, May 12 2014

a(0) inserted by Alois P. Heinz, Jan 21 2013

A122852 Row sums of number triangle A122851.

Original entry on oeis.org

1, 1, 2, 3, 6, 11, 24, 51, 122, 291, 756, 1979, 5526, 15627, 46496, 140451, 442194, 1414931, 4687212, 15785451, 54764846, 193129659, 698978136, 2570480147, 9672977706, 36967490691, 144232455524, 571177352091, 2304843053382, 9434493132011, 39289892366736

Offset: 0

Paul Barry, Sep 14 2006

Essentially the same as A072374. - R. J. Mathar, Jun 18 2008

Diagonal sums of A008279. - Paul Barry, Feb 11 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Jonathan Fang, Zachary Hamaker, and Justin Troyka, On pattern avoidance in matchings and involutions, arXiv:2009.00079 [math.CO], 2020. See Theorem 1.6 (b).
Guo-Niu Han, Hankel Continued fractions and Hankel determinants of the Euler numbers, arXiv:1906.00103 [math.CO], 2019. See p. 27.
Qiong Qiong Pan and Jiang Zeng, The gamma-coefficients of Branden's (p,q)-Eulerian polynomials and André permutations, arXiv:1910.01747 [math.CO], 2019.

Cf. A072374, A008279.

Cf. A357532, A357533, A357570.

Table[Sum[Binomial[n-k,k]*k!,{k,0,Floor[n/2]}],{n,0,20}] (* Vaclav Kotesovec, Feb 08 2014 *)

a(n) = sum(k=0, n, binomial(k,n-k)*(n-k)!); \\ Michel Marcus, Sep 02 2020

a(n) = Sum{k=0..n} C(k,n-k)*(n-k)!.
From Paul Barry, Feb 11 2009: (Start)
G.f.: 1/(1-x-x^2/(1-x^2/(1-x-2x^2/(1-2x^2/(1-x-3x^2/(1-3x^2/(1-x-4x^2/(1-4x^2/(1-... (continued fraction).
a(n) = Sum_{k=0..floor(n/2)} C(n-k,k)*k!. (End)
D-finite with recurrence -2*a(n) + 3*a(n-1) + (n-1)*a(n-2) + (-n+1)*a(n-3) = 0. - R. J. Mathar, Nov 15 2012. Proof in [Han 2019]
a(n) ~ sqrt(Pi) * exp(sqrt(n/2) - n/2 + 1/8) * n^((n+1)/2) / 2^(n/2+1) * (1 + 37/(48*sqrt(2*n))). - Vaclav Kotesovec, Feb 08 2014
a(n) = (a(n-1) + n * a(n-2) + 1)/2 for n > 1. - Seiichi Manyama, Nov 19 2022

More terms from Vaclav Kotesovec, Jun 04 2019

A049404 Triangle read by rows, the Bell transform of n!*binomial(2,n) (without column 0).

Original entry on oeis.org

1, 2, 1, 2, 6, 1, 0, 20, 12, 1, 0, 40, 80, 20, 1, 0, 40, 360, 220, 30, 1, 0, 0, 1120, 1680, 490, 42, 1, 0, 0, 2240, 9520, 5600, 952, 56, 1, 0, 0, 2240, 40320, 48720, 15120, 1680, 72, 1, 0, 0, 0, 123200, 332640, 184800, 35280, 2760, 90, 1, 0, 0, 0, 246400, 1786400

Offset: 1

Wolfdieter Lang

Previous name was: A triangle of numbers related to triangle A049324.
a(n,1) = A008279(2,n-1). a(n,m) =: S1(-2; n,m), a member of a sequence of lower triangular Jabotinsky matrices, including S1(1; n,m) = A008275 (signed Stirling first kind), S1(2; n,m) = A008297(n,m) (signed Lah numbers).
a(n,m) matrix is inverse to signed matrix ((-1)^(n-m))*A004747(n,m). The monic row polynomials E(n,x) := sum(a(n,m)*x^m,m=1..n), E(0,x) := 1 are exponential convolution polynomials (see A039692 for the definition and a Knuth reference).
For the definition of the Bell transform see A264428 and the link. - Peter Luschny, Jan 16 2016

			E.g. row polynomial E(3,x) = 2*x+6*x^2+x^3.
Triangle starts:
{1}
{2,  1}
{2,  6,  1}
{0, 20, 12, 1}

W. Lang, On generalizations of Stirling number triangles, J. Integer Seqs., Vol. 3 (2000), #00.2.4.
W. Lang, First 10 rows of the array and more. [From _Wolfdieter Lang_, Oct 17 2008]
Peter Luschny, The Bell transform

Row sums give A049425.

Cf. A004747, A049324.

rows = 11;
a[n_, m_] := BellY[n, m, Table[k! Binomial[2, k], {k, 0, rows}]];
Table[a[n, m], {n, 1, rows}, {m, 1, n}] // Flatten (* Jean-François Alcover, Jun 22 2018 *)

# uses[bell_matrix from A264428]
# Adds 1,0,0,0, ... as column 0 at the left side of the triangle.
bell_matrix(lambda n: factorial(n)*binomial(2, n), 8) # Peter Luschny, Jan 16 2016

a(n, m) = n!*A049324(n, m)/(m!*3^(n-m));
a(n, m) = (3*m-n+1)*a(n-1, m) + a(n-1, m-1), n >= m >= 1;
a(n, m) = 0, nE.g.f. for m-th column: ((x+x^2+(x^3)/3)^m)/m!.
a(n,m) = n!/(3^m * m!)*(Sum_{i=0..floor(m-n/3)} (-1)^i * binomial(m,i) * binomial(3*m-3*i,n)), 0 for empty sums. - Werner Schulte, Feb 20 2020

New name from Peter Luschny, Jan 16 2016

A071379 a(n) = (1/e) * Sum_{k>=0} ((k+4)!/k!)^(n-1)/k!.

Original entry on oeis.org

1, 1, 209, 163121, 326922081, 1346634725665, 9939316337679281, 119802044788535500753, 2205421644124274191535553, 58945667435045762187763602753, 2198513228897522394476415669503377, 110833342180980170285766876408530089329

Offset: 0

Karol A. Penson, May 22 2002

nonn

This is a Dobinski-type summation formula.
Terms quickly become gigantic: a(15) = 9142140479823239889945170786704021785456107245847570873873. a(n) appears in the process of ordering the n-th power of a product of fourth power of boson creation and fourth power of boson annihilation operators.
From Peter Luschny, Mar 27 2011: (Start)
Let B_{m}(x) = Sum_{j>=0} (exp(j!/(j-m)!*x-1)/j!) then a(n) = n! [x^n] Taylor(B_{4}(x)), where [x^n] denotes the coefficient of x^k in the Taylor series for B_{4}(x).
a(n) is row 4 of the square array representation of A090210. (End)

P. Blasiak, K. A. Penson and A. I. Solomon, The general boson normal ordering problem, arXiv:quant-ph/0402027, 2004.
P. Blasiak, K. A. Penson and A. I. Solomon, The general boson normal ordering problem, Phys. Lett. A 309 (2003) 198-205.
K. A. Penson, P. Blasiak, A. Horzela, A. I. Solomon and G. H. E. Duchamp, Laguerre-type derivatives: Dobinski relations and combinatorial identities, J. Math. Phys. 50, 083512 (2009).
M. Schork, On the combinatorics of normal ordering bosonic operators and deforming it, J. Phys. A 36 (2003) 4651-4665.

Cf. A000110, A020556 and A069223, when k+4 is replaced by k+1, k+2 or k+3 respectively.

Cf. A090210.

A071379 := proc(n) local r,s,i;
if n=0 then 1 else r := [seq(5,i=1..n-1)]; s := [seq(1,i=1..n-1)];
exp(-x)*24^(n-1)*hypergeom(r,s,x); round(evalf(subs(x=1,%),99)) fi end:
seq(A071379(n),n=0..10); # Peter Luschny, Mar 30 2011

a[n_] := Sum[FactorialPower[k, 4]^n/k!, {k, 4, Infinity}]/E; a[0] = 1; Array[a, 12, 0] (* Jean-François Alcover, Sep 01 2016 *)

default(realprecision, 500); for(n=0, 20, print1(if(n==0, 1, round(exp(-1)*sum(k=0, 500, ((k+4)!/k!)^(n-1)/k!))), ", ")) \\ G. C. Greubel, May 15 2018

a(n) = (1/e)*Sum_{k>=4} fallfac(k, 4)^n / k!, n >= 1, with fallfac(n, m) := A008279(n, m) (falling factorials). (From eq.(26) with r=4 of the Schork reference.)

E.g.f. with a(0) := 1: (1/e)*(Sum_{k>=4} e^(fallfac(k, 4)*x)/k! + 8/3). From top of p. 4656 with r=4 of the Schork reference.

a(0)=1 prepended by Alois P. Heinz, Aug 01 2016

If it is proved that A283153 and A071379 are the same, then the entries should be merged and A283153 recycled. - N. J. A. Sloane, Mar 06 2017

A090447 Triangle of partial products of binomials.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 3, 9, 9, 1, 4, 24, 96, 96, 1, 5, 50, 500, 2500, 2500, 1, 6, 90, 1800, 27000, 162000, 162000, 1, 7, 147, 5145, 180075, 3781575, 26471025, 26471025, 1, 8, 224, 12544, 878080, 49172480, 1376829440, 11014635520, 11014635520, 1, 9, 324

Offset: 0

Wolfdieter Lang, Dec 23 2003

			[1]; [1,1]; [1,2,2]; [1,3,9,9]; ...

W. Lang, First 10 rows.

Column sequences: A000027 (natural numbers), A006002, A090448-9.
Cf. A090450 (row sums), A090451 (alternating row sums).
Cf. A008949 (partial row sums in Pascal's triangle).
Cf. A000178, A007318, A008279.

a[n_, m_] := Product[Binomial[n, p], {p, 0, m}]; Table[a[n, m], {n, 0, 10}, {m, 0, n}] // Flatten (* Jean-François Alcover, Sep 01 2016 *)

a(n, m) = Product_{p=0..m} binomial(n, p), n>=m>=0, else 0. Partial row products in Pascal's triangle A007318.
a(n, m) = (Product_{p=0..m} fallfac(n, m-p))/superfac(m), n>=m>=0, else 0; with fallfac(n, m) := A008279(n, m) (falling factorials) and superfac(m) = A000178(m) (superfactorials).
a(n, m) = (Product_{p=0..m} (n-p)^(m-p))/superfac(m), n>=m>=0, with 0^0:=0, else 0.

A091752 Generalized Stirling2 array (-1,2)S2. Irregular triangle a(n, m) for n >= 1 and 2 <= m <= 2*n.

Original entry on oeis.org

1, 2, -2, 1, 40, -40, 20, -6, 1, 2240, -2240, 1120, -360, 80, -12, 1, 246400, -246400, 123200, -40320, 9520, -1680, 220, -20, 1, 44844800, -44844800, 22422400, -7392000, 1786400, -332640, 48720, -5600, 490, -30, 1, 12197785600, -12197785600, 6098892800, -2018016000, 493292800

Offset: 1

Wolfdieter Lang, Feb 27 2004

This a(n,m) array appears in the normal ordering formula ((1/x)*(d_x)^2)^n = Sum_{m=2..2*n} a(n,m)*x^(m-3*n)*(d_x)^m, n >= 1, with the derivative operator d_x := d/dx.
This is an extension of the generalized Stirling2 arrays S_{r,s}(n,k) (here k=m) considered for nonnegative r and s in the Blasiak et al. reference given in A078740. See also the Schork reference given there.
The sequence of row lengths for this array is [1,3,5,7,9,11,...] = A005408(n-1), n >= 1.
These generalized Stirling2 arrays have been treated in Carlitz's paper (with r = lambda + mu, s = mu), and a recurrence is given eq. (4). See the formula section for the present case mu = 2, lambda = -3, and Carlitz's a_{n,s-1} = a(n, s) (here s = m). - Wolfdieter Lang, Dec 16 2019
From Wolfdieter Lang and Werner Schulte, Jan 29 2020: (Start)
For the case of the (irregular) triangles (-1,s)S2, for s >= 1, W. Schulte conjectured that a(s; n, m) = T(s; (s+1)*n - m - 1, m - s), for n > = 1 and m = s, s+1, ..., s*n, with the row polynomials RT(s; n, x) = Sum_{m = 0..s*n} T(s; n, m)*x^m of the triangle T(s), n >= 0, defined by the Rodrigues-type formula exp(x^(s+1)/(s+1)) (d/dx)^n exp(-x^(s+1)/(s+1)) = (-1)^n*RT(s; n, x). Thus the rows of (-1,s)S2 are every (s+1)-th upwards antidiagonals of T(s), but with offset m = s instead of m = 0.
The proof of the conjecture follows by showing the Carlitz recurrence for (-1,s)S2, but with offset n = 0 and m = 0; that is, a(s; n+1, m+s) =: aHat(s; n, m) = Sum_{j = 0..s} binomial(s, j)*fallfac(s + m - j - (s+1)*n , s-j)*aHat(s; n-1, m-j), for n >= 0, m = 0, 1, ..., s*n, with aHat(s; 0, 0) = 1 and aHat(s; n, m) = 0 for m < 0 and m > s*n. The falling factorials are fallfac(x, n). With aHat(s; n, m) = T(s; (s+1)*n - m, m) this leads to a recurrence for T(s) which is equivalent to the recurrence for the row polynomials RT(s), namely RT(s; n, x) = Sum_{j=0 .. s} binomial(s, j)*x^j*fallfac(s - j - n, s - j)*RT(s; n - (s+1) + j, x), for n >= 1, and RT(s; 0, x) = 1. This, in turn, can be proved by induction over n >= 1 from the simpler recurrence for RT(s) obtained directly from the Rodrigues-type definition, namely RT(s; n, x) = x^s*RT(s; n-1, x) - (d/dx)RT(s; n-1, x), n >= 1,with RT(s; 0, x) = 1.
The e.g.f. of the triangle T(s), that is of the row polynomials {RT(s;n, x)}_{n>=0}, is E(s; t, x) = exp((x^(s+1) - (x - t)^(s+1))/(s+1)). This can be proved from the simple RT(s) recurrence, leading to (d/dt + d/dx)E(s; t, x) = x^s*E(s; t, x), with E(s; 0, x) = 1. After using E(s; t, x) = 1*exp(x^(s+1)/(s+1) + f(s; t, x)), with f(s; 0, x) = -x^(s+1)/(s+1), this becomes (d/dx - d/dt)f(s; t, x) = 0 meaning that f is a function of y = x - t, say, g(s; y) = -y^(s+1)/(s+1) because it has to become f(s; 0, x) for t = 0.
The explicit form for (-1,s)S2 is a(s; n, m) = (-1)^(n*s -m)*((s+1)*n - m-1)!/((s+1)^(n-1)*(n-1)!)*Sum_{j=0..floor((m-s)/(s+1))} (-1)^j* binomial(n-1, j)*binomial((s+1)*(n-1-j), m - s - (s+1)*j). One can prove the corresponding formula for T(s; n, m) by showing that it satisfies the T(s) recurrence T(s; n, m) = T(s; n-1, m-l) + (m+1)*T(s; n-1, m+1), for n >= 1, with T(s; 0, 0) = 1, and 0 for m < 0 or m > s*n.
The present entry is the instance s = 2, with the formulas given below. (End)

			Triangle starts:
{1},
{2, -2, 1},
{40, -40, 20, -6, 1},
{2240, -2240, 1120, -360, 80, -12, 1},
{246400, -246400, 123200, -40320, 9520, -1680, 220, -20, 1}.

Leonard Carlitz, On Arrays of Numbers, Am. J. Math., 54,4 (1932) 739-752.
Wolfdieter Lang, First 6 rows.

Column sequences for m=2..8 and n >= ceiling(m/2) are A052502(n-1), -A052502(n-1), A091535(n-1), -6*A091753(n), A091754(n), -12*A091755(n), A091756(n).

Cf. A078740, A331816.

w[n_, k_] := (Gamma[n-k/3] Gamma[1/3+n-k/3])/(Gamma[1/3-k/3] Gamma[-k/3]);
T[n_, k_] := 9^n Sum[(-1)^(k - j)  w[n, j]/((k - j)! j!), {j, 1, k}];
Table[Round[T[n,k]], {n,1,6}, {k, 2, 2 n}] (* Peter Luschny, Feb 05 2020 *)

a(n, m) = (((-1)^m)/m!)*Sum_{p=2..m} (-1)^p*binomial(m, p)*Product_{j=1..n} fallfac(p-3*(j-1), 2), n >= 1, 2 <= m <= 2*n, otherwise 0. From eq. (12) of the Blasiak et al. reference (see A078740) with r=-1, s=2, k=m.
Recurrence: a(n, m) = Sum_{p=0..2} binomial(2, p)*fallfac(-3*(n-1)+m-p, 2-p)*a(n-1, m-p), n >= 2, 2 <= m <= 2*n, a(1, 2) = 1, otherwise 0. Rewritten from eq. (19) of the Schork reference (see A078740) with r = -1, s = 2. fallfac(n, m) := A008279(n, m) (falling factorials triangle).
Recurrence (Carlitz): a(n m) = a(n-1, m-2) - 2*(3*n - (m +2))*a(n-1, m-1) + (3*n - (m + 3))*(3*n - (m + 2))*a(n-1, m), for n >= 2, m >= 1, and a(n, m) = 0 if m <= 1 or m > 2*n, and a(1, 2) = 1. - Wolfdieter Lang, Dec 16 2019
From Werner Schulte, Jan 29 2020: (Start)
a(n, m) = T(3*n - m - 1, m - 2), for n > = 1 and m = 2, 3, ..., 2*n, with the irregular triangle defined by (-1)^n*exp(x^3/3)*(d/dx)^n exp(-x^3/3) = RT(n, x) = Sum_{k=0..2*n} T(n, k)*x^k, for n >= 0. For T(n, k) see A331816.
The recurrence RT(n, x) = x^2*RT(n-1, x) - (d/dx)RT(n-1, x), n >= 1, with RT(0, x) = 1, implies the T recurrence T(n, k) = T(n-1, k-2) - (k+1)*T(n-1, k+1), for n >= 1, with T(0, 0) = 1, and T(n, m) = 0 for m < 0 and m > 2*n. Also, by induction over n: RT(n, x) = x^2*RT(n-1, x) - 2*(n-1)*x*RT(n-2, x) + (n-1)*(n-2)*RT(n-3, x), using the former recurrence and inserting the derivatives. This translates to an obvious further recurrence for the irregular triangle T. It is used in order to prove the Carlitz recurrence for the index shifted aHat(n, m) = a(n+1, m + 2).
The e.g.f. of the irregular triangle, that is of the row polynomials RT, is E(t, x) = exp((x^3 - (x - t)^3)/3). See the comment above for a proof (setting s=2 there).
The explicit form is a(n, m) = (-1)^m*(3*n - m - 1)!/(3^(n-1)*(n-1)!)*Sum_{j=0..floor((m-2)/3)} (-1)^j*binomial(n-1, j)*binomial(3*(n-1-j), m -2 - 3*j), for n >= 1, and 2 <= m <= 2*n.
(End)
T(n, k) = 9^n*Sum_{j=1..k} (-1)^(k-j)*w(n,j)/((k-j)!*j!) where w(n,k) = (Gamma(n-k/3)*Gamma((1-k)/3+n))/(Gamma((1-k)/3)*Gamma(-k/3)). - Peter Luschny, Feb 05 2020

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